Problem: Solve for $x$ : $6x^2 + 36x - 42 = 0$
Dividing both sides by $6$ gives: $ x^2 + {6}x {-7} = 0 $ The coefficient on the $x$ term is $6$ and the constant term is $-7$ , so we need to find two numbers that add up to $6$ and multiply to $-7$ The two numbers $7$ and $-1$ satisfy both conditions: $ {7} + {-1} = {6} $ $ {7} \times {-1} = {-7} $ $(x + {7}) (x {-1}) = 0$ Since the following equation is true we know that one or both quantities must equal zero. $(x + 7) (x -1) = 0$ $x + 7 = 0$ or $x - 1 = 0$ Thus, $x = -7$ and $x = 1$ are the solutions.